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JEE Main 2021
Complex Numbers
Complex Numbers
Medium

Question

If a > 0 and z = (1+i)2ai{{{{\left( {1 + i} \right)}^2}} \over {a - i}}, has magnitude 25\sqrt {{2 \over 5}} , then z\overline z is equal to :

Options

Solution

Key Concepts and Formulas

  • Complex Number Representation: A complex number can be represented as z=x+iyz = x + iy, where xx is the real part and yy is the imaginary part, and i=1i = \sqrt{-1}.
  • Modulus of a Complex Number: The modulus (or magnitude) of a complex number z=x+iyz = x + iy is given by z=x2+y2|z| = \sqrt{x^2 + y^2}.
  • Conjugate of a Complex Number: The conjugate of a complex number z=x+iyz = x + iy is given by z=xiy\overline{z} = x - iy.

Step-by-Step Solution

Step 1: Simplify the expression for zz.

We are given z=(1+i)2aiz = \frac{(1+i)^2}{a-i}. First, we simplify the numerator: (1+i)2=1+2i+i2=1+2i1=2i(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i. So, z=2iaiz = \frac{2i}{a-i}.

To express zz in the form x+iyx + iy, we multiply the numerator and denominator by the conjugate of the denominator, which is a+ia+i: z=2iaia+ia+i=2i(a+i)(ai)(a+i)=2ai+2i2a2i2=2ai2a2+1=2+2aia2+1z = \frac{2i}{a-i} \cdot \frac{a+i}{a+i} = \frac{2i(a+i)}{(a-i)(a+i)} = \frac{2ai + 2i^2}{a^2 - i^2} = \frac{2ai - 2}{a^2 + 1} = \frac{-2 + 2ai}{a^2 + 1}.

Separating the real and imaginary parts, we get: z=2a2+1+2aa2+1iz = \frac{-2}{a^2 + 1} + \frac{2a}{a^2 + 1}i.

Step 2: Use the given magnitude of zz to find aa.

We are given that z=25|z| = \sqrt{\frac{2}{5}}. Using the formula for the modulus, we have: z=(2a2+1)2+(2aa2+1)2=4(a2+1)2+4a2(a2+1)2=4+4a2(a2+1)2=4(1+a2)(a2+1)2=4a2+1=2a2+1|z| = \sqrt{\left(\frac{-2}{a^2 + 1}\right)^2 + \left(\frac{2a}{a^2 + 1}\right)^2} = \sqrt{\frac{4}{(a^2 + 1)^2} + \frac{4a^2}{(a^2 + 1)^2}} = \sqrt{\frac{4 + 4a^2}{(a^2 + 1)^2}} = \sqrt{\frac{4(1 + a^2)}{(a^2 + 1)^2}} = \sqrt{\frac{4}{a^2 + 1}} = \frac{2}{\sqrt{a^2 + 1}}.

Now we set this equal to the given magnitude: 2a2+1=25\frac{2}{\sqrt{a^2 + 1}} = \sqrt{\frac{2}{5}}. Squaring both sides, we get: 4a2+1=25\frac{4}{a^2 + 1} = \frac{2}{5}. Cross-multiplying, we have: 20=2(a2+1)20 = 2(a^2 + 1). Dividing by 2, we get: 10=a2+110 = a^2 + 1. So, a2=9a^2 = 9, which means a=±3a = \pm 3. Since a>0a > 0, we have a=3a = 3.

Step 3: Find the conjugate z\overline{z}.

Substitute a=3a = 3 into the expression for zz: z=232+1+2(3)32+1i=210+610i=15+35iz = \frac{-2}{3^2 + 1} + \frac{2(3)}{3^2 + 1}i = \frac{-2}{10} + \frac{6}{10}i = -\frac{1}{5} + \frac{3}{5}i.

Now, find the conjugate of zz: z=1535i\overline{z} = -\frac{1}{5} - \frac{3}{5}i.

Common Mistakes & Tips

  • Carefully handle the powers of ii. Remember that i2=1i^2 = -1.
  • When rationalizing the denominator, make sure to multiply both the numerator and the denominator by the conjugate of the denominator.
  • Pay attention to any given conditions, such as a>0a > 0, to choose the correct value.

Summary

We simplified the complex number zz, used its magnitude to find the value of aa, and then calculated the conjugate z\overline{z}. The final answer is z=1535i\overline{z} = -\frac{1}{5} - \frac{3}{5}i.

The final answer is \boxed{-\frac{1}{5} - \frac{3}{5}i}, which corresponds to option (B).

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